Bernstein inequality for bounded independent sums

Let $v_i=\mathbb{E}[X_i^2]$. If $v=0$, then every $v_i=0$ because $v_i\ge 0$. By [F6], $X_i^2=0$ almost surely for every $i$, so the finite sum is $0$ almost surely and the stated tail probability is $0$.

Suppose $v>0$. If $b=0$, every $X_i=0$ almost surely and $v=0$, a contradiction; hence $b>0$. For each fixed real $\theta$, $g_{\theta }(x)=e^{\theta x}$ is continuous: when $\theta \mathrm{\ne }0$, continuity of $x\mapsto \theta x$ follows from $\mathrm{\mid }\theta x-\theta y\mathrm{\mid }=\mathrm{\mid }\theta \mathrm{\mid }\mathrm{\mid }x-y\mathrm{\mid }$, and continuity of its composition with $e^x$ follows from [F10]; when $\theta =0$ it is constant. Thus $g_{\theta }$ is Borel: the Borel sets whose inverse images under $g_{\theta }$ are Borel form a sigma-algebra containing every open set, hence contain the sigma-algebra generated by opens (The Borel sigma-algebra of a topological space). For each Borel $B$, $\{g_{\theta }(X_i)\in B\}=X_i^{-1}(g_{\theta }^{-1}(B))$ is measurable. Now fix $i$ and $\lambda$ with $0<\mathrm{\mid }\lambda \mathrm{\mid }<3\mathrm{/}b$. For $N\ge 2$, the exponential series tail on $[-b,b]$ is uniformly bounded by $r_N={\sum }_{k>N}(\mathrm{\mid }\lambda \mathrm{\mid }b{)}^k\mathrm{/}k!\to 0$. Each partial sum at $X_i$ is integrable; the uniform bound and [F9] give $\mathrm{\mid }\mathbb{E}[e^{\lambda X_i}]-\mathbb{E}[{\sum }_{k=0}^N(\lambda X_i{)}^k\mathrm{/}k!]\mathrm{\mid }\le r_N$, so expectation passes to the series limit. Centering removes the linear term, and for every $k\ge 2$, $\mathrm{\mid }\mathbb{E}[X_i^k]\mathrm{\mid }\le b^{k-2}{v}_i$ by [F1] and monotonicity. Also $k!\ge 2\cdot 3^{k-2}$ for $k\ge 2$: equality holds at $k=2,3$, and each subsequent factorial factor is at least $3$. Hence the geometric-series bound and $1+x\le e^x$ from [F5] give $\mathbb{E}{e}^{\lambda X_i}\le 1+{\lambda }^2{v}_i\mathrm{/}[2(1-\mathrm{\mid }\lambda \mathrm{\mid }b\mathrm{/}3)]\le \exp ({\lambda }^2{v}_i\mathrm{/}[2(1-\mathrm{\mid }\lambda \mathrm{\mid }b\mathrm{/}3)])$.

Let $S={\sum }_{i=1}^n{X}_i$, a measurable real random variable by [F11]. By the addition formula [F10], $e^{\lambda S}={\prod }_i{e}^{\lambda X_i}$; independence [F8] and factorization [F2], applied to the nonnegative Borel functions established in step 1.2, give $\mathbb{E}{e}^{\lambda S}\le \exp ({\lambda }^{2}v\mathrm{/}[2(1-\lambda b\mathrm{/}3)])$ for $0<\lambda <3\mathrm{/}b$. Since $e^x$ is positive and increasing [F10], $S\ge t$ implies $e^{\lambda S}\ge e^{\lambda t}>0$, so [F3] and the reciprocal and addition formulas in [F10] give $\mathbb{P}(S\ge t)\le \exp (-\lambda t+{\lambda }^{2}v\mathrm{/}[2(1-\lambda b\mathrm{/}3)])$. Choose $\lambda =t\mathrm{/}(v+bt\mathrm{/}3)$; then $\lambda b\mathrm{/}3<1$ and direct simplification gives $\mathbb{P}(S\ge t)\le \exp (-t^2\mathrm{/}[2(v+bt\mathrm{/}3)])$.

For $0<\lambda <3\mathrm{/}b$, the bound in step 1.2 also holds at parameter $-\lambda$. By [F10], $e^{-\lambda S}={\prod }_i{e}^{-\lambda X_i}$; the original variables remain independent by [F8], and [F2] factors these nonnegative Borel functions, giving $\mathbb{E}{e}^{-\lambda S}\le \exp ({\lambda }^{2}v\mathrm{/}[2(1-\lambda b\mathrm{/}3)])$. Since $S\le -t$ implies $e^{-\lambda S}\ge e^{\lambda t}>0$, Markov [F3] and the same optimizing choice as in step 2.1 yield $\mathbb{P}(S\le -t)\le \exp (-t^2\mathrm{/}[2(v+bt\mathrm{/}3)])$. By [F7], the two one-sided bounds imply the displayed two-sided bound without asserting independence of the transformed variables $-X_i$.